1. PHL424: Rutherford scattering discovery of nucleus

2. PHL424: Rutherford scattering discovery of nucleus
1909: Rutherford, Geiger and Marsden studied in Manchester
the scattering of α-particles on thin gold foils.
Aim: from the angular distribution of the scattered
α-particles they wanted to gain information on
the structure of the scattering center.
Experimental set-up: Ra-source with Ekin(α) = 4.78 MeV
thin Au-foils (Z = 79, d = 2000 atomic layers)
detection of scattered α‘s with ZnS scintillator
1908: Nobel price
for chemistry
Ernest Rutherford
( )

3. Rutherford scattering discovery of nucleus
observation:
𝑒𝑣𝑒𝑛𝑡 𝑟𝑎𝑡𝑒 ~ 1 𝑠𝑖𝑛 4 𝜃 2
scattering on a point-like atomic nucleus

4. Rutherford scattering discovery of nucleus
Kinematic of elastic scattered α-particles
energy and momentum conservation
α-particles on electrons (Thomson model)
𝑚 𝑒 𝑚 𝛼 ≈ 10 −4
max. momentum transfer Δp ~ 10-4·pi
only small scattering angles θ ~ 00
α-particles on Au-nuclei (Rutherford model)
max. momentum transfer Δp ~ 2·pi
scattering angles up to θmax ~ 1800
(backscattered α-particles)
𝑚 𝐴𝑢−197 𝑚 𝛼 ≈50
mα = 4 GeV/c2
me = MeV/c2
mAU-197 = 197 GeV/c2

5. Elastic scattering 𝑝 𝑎′ 𝑝 𝑎′ 𝑞 = 𝑝 𝑎′ − 𝑝 𝑎 𝑝 𝑎 𝑝 𝑎 θ
In an elastic process
a + b → a’ + b’
the same particles are present both before and after the scattering, i.e. the initial and the final state are identical (including quantum numbers) up to momenta and energy.
incoming particle:
projectile
target nucleus
The target b remains in its ground state, absorbing merely the recoil momentum and hence changing its kinetic energy.
The scattering angle and the energy of the projectile a and the recoil energy and energy of the target b are unambiguously correlated.
scattered outgoing particle a’
𝑝 𝑎′
𝑝 𝑎′
𝑞 = 𝑝 𝑎′ − 𝑝 𝑎
beam particle a
scattering angle θ θ
𝑝 𝑎
𝑝 𝑎
target nucleus b

6. Rutherford scattering
𝑉 𝑟 ~ 𝑧∙𝑍∙ 𝑒 2 𝑟
In the repulsive Coulomb potential the α-particle experiences a momentum change
∆ 𝑞 = 𝑝 𝑓 − 𝑝 𝑖
∆𝑞=2∙𝑚∙𝑣∙𝑠𝑖𝑛 𝜃 2
∆𝑞= 𝐹 ∆𝑞 ∙𝑑𝑡 = −∞ +∞ 1 4𝜋 𝜀 0 ∙ 𝑧∙𝑍∙ 𝑒 2 𝑟 2 ∙𝑐𝑜𝑠𝜙∙𝑑𝑡
∆𝑞= − 𝜋−𝜃 /2 𝜋−𝜃 /2 1 4𝜋 𝜀 0 𝑧𝑍 𝑒 2 𝑣∙𝑏 ∙𝑐𝑜𝑠𝜙∙𝑑𝜙= 1 4𝜋 𝜀 0 𝑧𝑍 𝑒 2 𝑣∙𝑏 ∙2∙𝑐𝑜𝑠 𝜃 2
angular momentum:
𝐿 = 𝑟 ×𝑚∙ 𝑣 =𝑚∙𝑣∙𝑏=𝑚∙ 𝑑𝜙 𝑑𝑡 ∙ 𝑟 2
relation between impact parameter b and scattering angle θ:
position
r(t), ϕ(t)
𝑡𝑎𝑛 𝜃 2 = 1 4𝜋 𝜀 0 𝑧∙𝑍∙ 𝑒 2 𝑚∙ 𝑣 2 ∙ 1 𝑏 ϕ

7. Rutherford scattering scattering parameters
impact parameter:
𝑏=𝑎∙𝑐𝑜𝑡 𝜃 𝑐𝑚 2
distance of closest approach:
𝐷=𝑎∙ 𝑠𝑖𝑛 −1 𝜃 𝑐𝑚 2 +1
orbital angular momentum: D
ℓ= 𝑘 ∞ ∙𝑏=𝜂∙𝑐𝑜𝑡 𝜃 𝑐𝑚 2
θcm
half distance of closest approach
in a head-on collision (θcm=1800):
𝑎= 0.72∙ 𝑍 1 𝑍 2 𝑇 𝑙𝑎𝑏 ∙ 𝐴 1 + 𝐴 2 𝐴 𝑓𝑚 b D
asymptotic wave number:
𝑘 ∞ =0.219∙ 𝐴 2 𝐴 1 + 𝐴 2 ∙ 𝐴 1 ∙ 𝑇 𝑙𝑎𝑏 𝑓𝑚 −1
Sommerfeld parameter:
𝜂= 𝑘 ∞ ∙𝑎=0.157∙ 𝑍 1 𝑍 2 ∙ 𝐴 1 𝑇 𝑙𝑎𝑏
center of mass system

8. Cross section
The cross section gives the probability of a reaction between the two colliding particles
Consider an idealized experiment:
we bombard the target with a monoenergetic beam of point-like particles a with a velocity va.
a thin target of thickness d and a total area A with Nb scattering centers b and with a particle density nb.
each target particle has a cross-sectional area σb, which we have to find by experiment!
→ Some beam particles are scattered by the scattering centers of the target, i.e. they are deflected from their original trajectory. The frequency of this process is a measure of the cross section area of the scattered particles σb.

9. Rutherford cross section
Particles from the ring defined by the impact parameter b and b+db scatter between angle θ and θ+dθ
𝑗∙2𝜋∙𝑏∙𝑑𝑏=𝑗∙2𝜋∙𝑠𝑖𝑛𝜃∙ 𝑑𝜎 𝑑Ω
𝑑𝜎 𝑑Ω = 𝑏 𝑠𝑖𝑛𝜃 𝑑𝑏 𝑑𝜃
𝑏=𝑎∙𝑐𝑜𝑡 𝜃 2
impact parameter:
ring area:
2πb db
𝑑𝑏 𝑑𝜃 = 𝑎 2 ∙ −𝑠𝑖𝑛 𝜃 2 ∙𝑠𝑖𝑛 𝜃 2 −𝑐𝑜𝑠 𝜃 2 ∙𝑐𝑜𝑠 𝜃 2 𝑠𝑖𝑛 2 𝜃 2 = 𝑎 2 ∙ 1 𝑠𝑖𝑛 2 𝜃 2
solid angle:
2π sinθ dθ
𝑑𝜎 𝑑Ω =𝑎∙ 𝑐𝑜𝑠 𝜃 2 𝑠𝑖𝑛 𝜃 2 ∙ 1 2∙𝑐𝑜𝑠 𝜃 2 ∙𝑠𝑖𝑛 𝜃 2 ∙ 𝑎 2∙ 𝑠𝑖𝑛 2 𝜃 2
𝑑𝜎 𝑑Ω = 𝑎 2 4 ∙ 𝑠𝑖𝑛 −4 𝜃 2
an atomic nucleus exist

10. Annular gas-filled parallel-plate avalanche counter (PPAC)
V0 ~ 500 V
p = 5-10 Torr
gap ~ 3 mm (anode-cathode)
122Sn
58Ni beam
Δt ~ tanϑ i o i
delay line
Δtime →
A. Jhingan detector laboratory at IUAC

11. Experimental set-up at IUAC
TARGET
CLOVER DETECTORS
@ backward angles
forward angle

12. Elastic scattering and nuclear radius
2 4 𝐻𝑒 𝑃𝑏
θ1/4 = 600, Eα = 30 MeV
→ Rint = 12.0 [fm]
𝑅 𝑖𝑛𝑡 ≅ 𝑅 𝐻𝑒 + 𝑅 𝑃𝑏 +3𝑓𝑚
𝐷=𝑎∙ 𝑠𝑖𝑛 −1 𝜃 𝑐𝑚 2 +1
𝑎= 0.72∙ 𝑍 1 𝑍 2 𝑇 𝑙𝑎𝑏 ∙ 𝐴 1 + 𝐴 2 𝐴 𝑓𝑚
R.M. Eisenberg and C.E. Porter, Rev. Mod. Phys. 33, 190 (1961)

13. Elastic scattering and nuclear radius
Nuclear interaction radius: (distance of closest approach)
𝑅 𝑖𝑛𝑡 =𝐷=𝑎∙ 𝑠𝑖𝑛 −1 𝜃 1/
𝑅 𝑖𝑛𝑡 = 𝐶 𝑝 + 𝐶 𝑡 +4.49− 𝐶 𝑝 + 𝐶 𝑡 𝑓𝑚
𝐶 𝑖 = 𝑅 𝑖 ∙ 1− 𝑅 𝑖 − 𝑓𝑚
𝑅 𝑖 =1.28∙ 𝐴 𝑖 1/3 − ∙ 𝐴 𝑖 −1/ 𝑓𝑚
Nuclear density distributions at the nuclear interaction radius
θ1/4 = 600 → Rint = 13.4 [fm]
→ ℓgr = 152 [ħ]
J.R. Birkelund et al., Phys.Rev.C13 (1976), 133